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PROFESSOR: Levinson's theorem,
in terms of derivations,

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that we do in this
course, this is probably

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the most subtle derivation
of the semester.

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It's not difficult, but
it's kind of interesting

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and a little subtle.

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And it's curious, because
it relates to things that

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seem to be fairly unrelated.

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But the key thing
that one has to do

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is you have to use something.

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How all of the
sudden are you going

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to relate phase shifts
to bound states?

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The one thing you
have to imagine

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is that if you have
a potential and you

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have states of a potential,
if the potential changes,

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the states change.

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But here comes
something very nice.

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They never appear or disappear.

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And this is something probably
you haven't thought about this

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all that much.

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Because you had, for example,
a square, finite square well.

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If you made this more deep,
you've got more bound states.

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If you made it less deep,
the bound states disappear.

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What does it mean the
bound states disappear?

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Nothing, really can disappear.

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What really is happening,
if you have the bounds,

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the square well.

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You have a couple of
bound states, say.

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But then you have an infinity
of scatterings states.

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And as you make this
potential less shallow,

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the last bound state
is approaching here.

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And at one state,
it changes identity

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and becomes a scatterings
state, but it never gets lost.

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And how, when you make
this deeper and deeper

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you get more state, is
the scattering state

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suddenly borrowing, lending
you a state that goes down?

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The states never get
lost or disappear.

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And you will, say, how
could you demonstrate that?

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That sounds like
science fiction,

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because, well, there's
infinitely many states here.

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How do you know it borrowed one?

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Well, you can do it by putting
it in a very large box.

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And then the states
here are going

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to be finitely
countable and discrete.

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And then you can
track and see indeed

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how the states become bound.

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So you'd never lose
or gain states.

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And that's a very, very powerful
statement about quantum states

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in a system.

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So this is what
we're going to need

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to prove Levinson's theorem.

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So Levinson theorem theorem--

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so it relates a number of
bound states of the potential

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to the excursion
of the phase shift.

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So let's state it completely.

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It relates the number N of
bound states of the potential

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to the excursion, excursion of
the phase from E equals 0 to E

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equals infinity.

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So in other words, it says
that N is 1 over pi delta of 0

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minus delta of
infinity, a number

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of bound states
of your potential

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is predicted by the behavior of
the phase shift of scattering.

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So how do we do this?

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This is what we want to prove.

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So consider, again, our
potential of range R and 0

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here and here is x.

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And I want you to be
able to count states,

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but discovering states are
a continuous set of states.

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So in order to
count states, we're

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going to put a
wall here, as well,

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at some big distance L
much bigger than A than R.

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And we're going, therefore--

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now the states are
going to be quantized.

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They're never going be
quite scatterings states.

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They're going to look
like scatterings states.

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But they're precisely
in the way that they

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vanished at this point.

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Now you would say, OK, that's
already a little dangerous

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to be.

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Oh, you've changed the
problem a little, yes.

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But we're going
to do the analysis

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and see if the result depends
on L. If it doesn't depend on L

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and L is very large, we'll
take the limit this L goes

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to infinity.

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And we claim, we have answer.

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So we argue that L is
a regulator, regulator

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to avoid a continuum,
continuum of states

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to avoid that
continuum of states.

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All right, so let's
begin to count.

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To count this thing,
we start with the case

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where there is no
potential again.

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And why is that?

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Because we're going
to try to compare

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the situation with no
potential to the situation

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with potential.

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So imagine let V identical
is 0, no potential

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and consider positive
energy states.

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These are the only
states that exist.

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There are no bound states,
because the potential is 0.

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Well, the solutions
were found before, we

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mentioned that these are what we
call phi effects or sine of kx.

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But now we require that
phi of L is equal to 0,

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because we do have
the wall there.

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And therefore, si of kl must
be 0 and kl must be n pi and n

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is 1, 2 to infinity.

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You know we manage with the
wall to discretize this state,

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because the whole world is now
an infinite, a very big box,

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not infinite, but very big.

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So you've discrete the state.

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The separations are
microelectron volts,

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but they are discrete.

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You can count them.

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And then with this
state over here,

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we think of counting them.

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And you say, well, I
can count them with n.

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So if I imagine the
k line from 0 to 50,

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the other states are
over, all the values of k.

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And they could say,
well, I even want

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to see how many states there
are in a little element dk.

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And for the that you
would have that dk taken

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a differential here
is dn times pi.

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So of the number of states
that there are in dk, dn--

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let me right here--

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dn equals l over pi dk is
the number of positive energy

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states in dk.

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In the range dk, in
the range of momentum,

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dk that little interval, there
are that many positive energy

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eigenstates.

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So far, so good.

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So now let's consider
the real case.

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Repeat for the case
there is some potential.

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Well, you would say, well,
I don't know how to count.

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I have to solve the problem
of when the potential makes

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a difference.

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But no, you do
know how to count.

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So repeat for V
different from 0.

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This time we have a
solution for x greater

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than R. We know the solution.

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This is our universal
solution with the phase.

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So there you have the si
effect is e to i delta sine

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kx plus delta of k.

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That's a solution.

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Yes, you have the
solution always.

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You just don't
know what delta is.

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But you don't know what delta--

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you don't need to know what
delta is to prove the theorem,

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You just need to
know it's there.

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So here it goes.

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And this time the wall will
also do the same thing.

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We'll demand the si of x
vanishes for x equal L.

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So this time we get that kx--

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no kL plus delta of k is
equal to some other number n

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prime times pi multiple of pi.

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This phase-- this total phase
has to be a multiple of pi.

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And what is n prime?

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I don't know what is n prime?

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It is some integers.

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I don't know whether it starts
from 1, 2, 3 or from 100

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or whatever.

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The only thing we
care is that, again,

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taking a little
differential, you

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have dk times L
plus d delta, dk,

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times dk is equal to
the dn prime times pi.

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We take an infinitesimal version
of this equation, which again

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tells me how many positive--
all these states are

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positive energy states.

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They have k.

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So all these are
positive energy states.

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So from this equation, we get
that dn prime is equal to L

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over pi dk plus 1 over pi
d delta dk times dk, which

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gives me if I know the dk,
again, the number of states

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that you have in
that range of k,

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You see it's like
momentum is now quantized.

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So for any little
range of momentum,

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you can tell how many
states there are.

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And here it is how many states
there are, positive energy

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states, positive.

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This is the number of positive
energy states in dk with V

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different from 0, here
is with V equal 0.

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